(related to Proposition: Definition of Integers)

Let \((a,b),(c,d)\) be ordered pairs of natural numbers. We will prove that the relation defined using addition of natural numbers. \[(a,b)\sim (c,d)\quad\Longleftrightarrow\quad\begin{cases}(a+h,b+h)=(c,d)& or\(a,b)=(c+h,d+h).\end{cases}]`

is an equivalence relation. In other words, the integers as equivalence classes \([a,b]:=\{(c,d),~(c,d)\sim (a,b)\}\) are well-defined.

\(( i )\) Reflexivity \((a,b)\sim (a,b)\)

Clearly, \((a+0,b+0)=(a,b)\).

\(( ii )\) Symmetry \((a,b)\sim (c,d)\Leftrightarrow(c,d)\sim(a,b)\)

If \((a+h,b+h)=(c,d)\), then \((c,d)=(a+h,b+h)\), otherwise if \((a,b)=(c+h,d+h)\), then \((c+h,d+h)=(a,b)\).

\(( iii )\) Transitivity: If \((a,b)=(c,d)\) and \((c,d)\sim (e,f)\), then \((a,b)\sim (e,f)\)

For symmetry reasons \(( ii) \), it is sufficient to consider only this case: By hypothesis, \((a+h_1,b+h_2)=(c,d)\) and \((c+h_2,d+h_2)=(e,f)\) for some \(h_1,h_2\in\mathbb N\). Therefore \((a+(h_1+h_2)),b+(h_1+h_2))=(e,f)\).

Thank you to the contributors under CC BY-SA 4.0!




  1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013